`(dy)/(dx)=(1+tan\ x)^2sec^2 2x`

We need to find:

`y=int(1+tan\ 2x)^2sec^2\ 2x\ dx`

Put `u=1+tan\ 2x`, then `du=2\ sec^2 2x\ dx`

`y = int(1+tan\ 2x)^2 sec^2 2x\ dx`

`=1/2intu^2du`

`=(u^3)/6+K`

`=((1+tan\ 2x)^3)/6+K`

The curve passes through `(2, 1)`.

This means when `x = 2`, `y = 1`. Substituting gives:

`1=((1+tan\ 2(2))^3)/6+K`

`1=((1+tan\ 4)^3)/6+K`

`1=1.674539+K`

This gives:

`K~~-0.675`

So we finally have the required equation for y:

`y=((1+tan\ 2x)^3)/6-0.675`

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